Enumeration under group action: Unsolved graphical enumeration problems, IV

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<ul><li><p>JOURNAL OF COMBINATORIAL THEORY 8, 1--11 (1970) </p><p>Enumeration under Group Action: Unsolved Graphical Enumeration Problems, IV* </p><p>FRANK HARARY** </p><p>Department of Mathematics, University of Michigan, Ann Arbor, Michigan 48104 </p><p>Received November 29, 1968 </p><p>ABSTRACT </p><p>This review article presents three methods for solving enumeration problems which can be construed as the determination of the number of orbits of an appropriate permutation group. Such a group must be constructed in accordance with the idio- syncrasies of the configurations to be counted and the equivalence relation on them. Thus three different binary operations on permutation groups, the sum, product, and power group, are defined and the structure of each is investigated. Applications of the corresponding theorems for enumeration under group action are provided. We conclude with a table of 27 current unsolved problems in graphical enumeration. </p><p>1. INTRODUCTION </p><p>There is a considerable interest in graphical enumeration problems, but apparent ly a major ity of otherwise bril l iant graph theorists cannot count. Why ? Perhaps it is because the combinator ia l aspects of enumerat ion are a bit more subtle than the conventional methods of p roo f used in structural graph theory. It may be the presence of permutat ion groups and the concomitant algebraic formulas and viewpoints. A closely related and somewhat equivalent approach using generalized M6bius inversion formulas (see Rota [13]), for enumeration of configurations by studying associated lattices, does not seem to make counting problems any more palatable to the world of graph theorists. </p><p>In an effort to make some of the standard enumeration methods available, we develop in this article three enumeration theorems, already existing in the literature, in a unified presentation. Each of these is asso- </p><p>* Presented at the Yale University Conference on Combinatorial Theory in honor of Professor Oystein Ore (May, 1968). </p><p>** Research supported in part by a grant from the U. S. Air Force Office of Scientific Research. This paper was prepared with the considerable assistance of Paul Stockmeyer and Dennis Geller. </p><p>1 9 1970 by Academic Press, Inc. </p></li><li><p>2 HARARY </p><p>ciated with an operation on permutation groups: the sum, product, and power of two groups. These theorems are illustrated by finding the number of necklaces of certain types. </p><p>For solving new enumeration problems, surely additional combinations of permutation groups will need to be devised. One such operation, for example, is that discovered by Palmer and Robinson [10], called the matrix group of two groups. A list of 27 unsolved combinatorial enumera- tion problems involving graphs and related structures concludes this article. </p><p>2. THE THEOREMS OF BURNSIDE AND POLYA </p><p>The fundamental result in the theory of enumeration under group action is the theorem of Burnside [2, p. 191], which counts the number of orbits of a permutation group. Most graphical enumeration problems are con- cerned only with the number of non-isomorphic graphs of a given type. Thus a typical solution consists of first finding a permutation group whose orbits are exactly the isomorphism classes of graphs under consideration, and then applying an appropriate generalization of Burnside's Theorem. </p><p>Let A be a permutation group acting on a set X. For each ~ ~ A, we definejk(~) as the number of cycles of length k in the disjoint cycle decom- position of ~. In particular, jl(~) is the number of elements of X fixed by a. </p><p>BURNSIDE'S THEOREM. group A is given by </p><p>The number N(A) of orbits of the permutation </p><p>1 N(A) = I A -~ ~ .~(cx). (1) </p><p>~x~A </p><p>As a typical application of this theorem, we will count the number of non-equivalent necklaces containing n beads, each of which has one of m colors. Two necklaces are considered equivalent if one can be rotated or reflected onto the other. Casting this problem in more mathematical terms, we will count the number of equivalence classes of functions whose domain is the set of n places on the framework of a necklace and whose range is the set of m colors. Two functions f and g are then equivalent if there exists a permutation ~ in the dihedral group D~, acting on the places in the framework, such thatf(x) = g(ax). We can now define a permuta- tion group /~i induced by D~ and acting on these functions f by Of(x) = f(~c) where ~ ~ D, and (~ is the corresponding element o f / )~ . Of course D, also depends on the number m of colors. Then the equiva- </p></li><li><p>UNSOLVED GRAPHICAL ENUMERATION PROBLEMS, IV 3 </p><p>lence classes of functions are exactly the orbits o f /~n. By Burnside's Theorem, the number N(/~n) of necklaces on n places is given by </p><p>1 = Z </p><p>c~D n </p><p>For a specific example, take n = 5, m = 2. Thus we are counting necklaces with 5 beads of two colors, say black and white. The identity element in fi 5 (which of course depends also on the value m = 2) fixes all 2 ~ functions, the four permutations that correspond to rotations of the framework each fix only 2 functions, and the five permutations correspond- ing to reflections each fix 2 3 functions. Thus the number N(/)s) of neck- laces with 5 places and 2 colors is given by </p><p>N(/~5) = 1 (25 q_ 4 9 21 -~- 5 9 2 3) = 8. </p><p>These 8 necklaces are illustrated in Figure 1, where the white beads are the points with open dots. </p><p>FIGURE 1 </p><p>The major weakness of Burnside's Theorem is that it does not easily yield the number of necklaces with a given number of beads of each color; more generally, it does not give the number of "configurations" with a given "weight." A powerful generalization which handles such questions was developed by P61ya [11] by exploiting generating functions. </p><p>We first define the cycle index Z(A) of a permutation group A of degree d as the polynomial in d variables al ..... aa given by </p><p>1 a ,7 ~'~(~) Z(A)---- f A--S Z 9 (2) </p><p>a~A k=l </p><p>For any permutation ~ ~ A, the numbers.]k = Jk(o0 satisfy </p><p>Jx ~- 2j2 + "'" q- dj, = d, (3) </p></li><li><p>4 HARARY </p><p>and constitute a partition of the integer d. It is useful to employ the vector notation </p><p>(J) = 01 ..... J~) </p><p>to express the cycle structure of a. The general form of a counting problem to which P61ya's Theorem </p><p>applies is similar to that for Burnside's Theorem. Not only is there given a domain D and a range R, but in addition a weight function w defined on R. In a typical application, w might assign to each r e R an ordered pair w(r) = (wlr, w2r) of non-negative integers. This leads to a statement of P61ya's Theorem displaying just two variables x and y, a convenient special case. Once again, the objects to be counted are functions from D into R. A group A acting on D is given, and two functions f and g are called A-equivalent if there exists e~ e A such that f(d) = g(ad). Thus A induces a permutation group A on R D whose orbits are the equivalence classes of functions. </p><p>We assign another kind of weight W(f ) to each funct ion fe R D by </p><p>W(f) = 1-I x~Ol*(a)Y w~*(a)" (4) deD </p><p>Each function in a given orbit of A then has the same weight, so the weight of an orbit can be defined as the weight of any function in it. </p><p>Suppose the number of range elements of weight (m, n) is c . . . . and the number of orbits in R ~ with weight xmy ~ is C~.,n 9 Thefigure counting series is </p><p>c(x, y) = ~, cm,,x~y"; (5) </p><p>it enumerates the elements of R by weight. Similarly the configuration counting series </p><p>C(x, y) = E Cm,',xmY" (6) </p><p>is the generating function for equivalence classes. Finally, for any function h(x, y), we define Z(A; h(x, y)) as the function </p><p>obtained from Z(A) on replacing each variable ai by h(x ~, yi). </p><p>P6LYA'S THEOREM. The configuration counting series is obtained by substituting the figure counting series into the cycle index of the configura- tion group, </p><p>C(x, y) = Z(A; c(x, y)). (7) </p><p>We return to the necklace example, with D the set of places on the J necklace framework and R = {black, white}. Let black be assigned the </p></li><li><p>UNSOLVED GRAPHICAL ENUMERATION PROBLEMS, IV 5 </p><p>weight (1,0) and white the weight (0,1), so that c(x, y )= x + y. The group D 5 has cycle index </p><p>Z(D5 ) = 1 (aln + 4a 5 q_ 5ala2~), </p><p>so by P61ya's Theorem, </p><p>C(x, y) = Z(Ds, x + y) </p><p>1 = 10 ((x + y)5 + 4(x 5 -k yS) + 5(x + y)(x 2 -}- y2)=) </p><p>= x 5 + x4y + 2xay2 + 2x~ya + xy4 + yS. </p><p>These coefficients are verified in Figure 1. We thus know not only that there are 8 necklaces, but also the number of necklaces with each color combination. </p><p>Graphs withp points and q lines can be enumerated in the same manner. Let D = V (m be the set of (~) pairs of points, and R the set {yes, no}, representing the presence or absence of a line joining a given pair of points, Now consider the one-variable form of P61ya's Theorem, C(x) = Z(A; e(x)), and assign "yes" the weight 1 and "no" the weight 0, so that c(x) = 1 + x. The group acting on the domain V (~) is the symmetric pair group S~ ~), which is induced by the symmetric group S~ acting on the set V of p points. P61ya first showed (see [3, 8]) that the cycle index of this group is given by </p><p>1 p ! ['/~] (~j) I'aku2~ ] Z(S(~2)) = p l . 1-IP=I Jlc ! kJk k=lH [ --k--l'lJ2k </p><p>(8) t -1,,1 E ,,1 ) H -,.+,-'J"+' II 17 . . . . , . , , k=o k=l l</p></li><li><p>6 HARARY </p><p>3. OPERATIONS ON PERMUTATION GROUPS AND THE INDUCED ENUMERATION THEOREMS </p><p>Further generalizations can be obtained by considering binary opera- tions on permutation groups. In fact, it seems to be a meta-theorem that for every binary operation on permutation groups there is a corresponding enumeration theorem. We now examine three such operations. </p><p>Let A be a permutation group of order m = [ A ] and degree d acting on the set X = {x I , x 2 . . . . . xd} , and let B be another permutation group of order n = IB[ and degree e acting on the set Y={y l ,y2 ..... Ye}- Their sum A + B is a permutation group which acts on the disjoint union X w Y and whose elements are all the ordered pairs, written as the juxta- position ~/3, of permutations o~ in A and/3 in B. Any element z of X tj Y is permuted by o~fl according to the rule: </p><p>I O~Z~ Z ~ X, o~5(z) = ~/3z, z ~ r . </p><p>The product A x B of A and B is a permutation group which acts on the set X x Y and whose permutations are all the ordered pairs, written (o~, fi), of permutations a in A and fl in B. The element (x, y) of X x Y is permuted by (~,/3) as expected: </p><p>(~, f l ) (x, y ) = (o~x, f ly). </p><p>The power group "B to the A" acts on yx , the set of all functions from X into I1, and is denoted by B A. We will always assume that the power group acts on more than one function. For each pair of permutations </p><p>in A and fl in B there is a unique permutation, written (~; fl) in B A. We specify the action of (~; fl) on any function f in yx by the following equation which gives the image of each x ~ X under the function (cq f l ) f : </p><p>((cq f l) f ) (x ) = f l f (o~x). (14.4) </p><p>We summarize this information in the following table: </p><p>sum product power </p><p>group A B A + B A x B B a objects X Y W u Y X Y yx order m n mn mn mn degree d e d -b e de e a </p></li><li><p>UNSOLVED GRAPHICAL ENUMERATION PROBLEMS, IV 7 </p><p>We remark in passing that the sum, product, and power group are all isomorphic as abstract groups, although of course they are different as permutation groups. For each of these three binary operations, there is a corresponding enumeration theorem. The first of these appeared in P61ya's original paper [11]. </p><p>SUM GROUP ENUMERATION THEOREM. The cycle index of the sum of two groups is the product of the cycle indices of the groups, </p><p>z(A + B) = z(A) z(B). (10) </p><p>As an example, consider the number of pairs of necklaces a person could wear, one of length 4 and one of length 5, both with black and white beads. We already know that </p><p>Z(D 5 , x y) = x 5 q- x4y + 2x3y 2 q-2x2y 3 q- xy 4 q- yS. </p><p>We find similarly that </p><p>Z(D4, x -~ y) = x 4 -~- x3y -~- 2x2y 2 + xy z + ya. </p><p>Then the number of such pairs of necklaces is counted by </p><p>Z(D4 q- Ds, x q- y) = Z(D4, x q- y) Z(D5, x q- y) </p><p>= x 9 + 2xSy + 5xTy 2 + 7xay 3 -I- 9xSy 4 </p><p>_5 9x4y5 + 7x3ye q_ 5x2y7 + 2xyS q_ yD. </p><p>The theorem corresponding to the product of two groups is somewhat more complex [4]. </p><p>PRODUCT GROUP ENUMERATION THEOREM. The cycle index of the product A B is given by </p><p>1 a" ,.(cO.(~) am(r, s) Z(A B) = I A t IB I ~' ]-I tr.8,,~ ~, (11) </p><p>We illustrate this theorem by counting bicolored graphs with 2 points of one color (white) and 3 points of another (black). Let 2" be a set of 2 points and Y a disjoint set of 3 points. The domain then is 2" Y and the permutation group acting on it is the product $2 $3 9 Then by (11) we have, </p><p>Z(S3 5'2) = 1 (a e + 3a12a2 ~ + 2aa2 + 4a3 _q_ 2ae). IZ , , </p></li><li><p>8 HARARY </p><p>As in counting graphs, the figure series is again 1%- x. After substituting 1%- x i for ai, we obtain the generating function for the bicolored graphs, with 2 points of one color and 3 points of another, enumerated by number of lines: </p><p>Z(S3247 1 +x ~+3x 3%-3x ~%-x 5%-x 6. </p><p>The corresponding graphs are shown in Figure 2. </p><p>o ~ </p><p>o o </p><p>FIGURE 2 </p><p>All the examples considered up to now have been cases of counting functions when a permutation group acted on the domain. Using the notation of power groups, we can express the group that acted on the functions in R ~ as E A, where A is the group acting on the domain ele- ments and E is the identity group acting on the range. It is natural to consider what happens when non-trivial groups act on both the domain and the range. </p><p>The next result was first found by de Bruijn [1]; the equation (12) below gives the version which appears in [9], where it was discovered that the power group is involved. </p><p>POWER GROUP ENUMERATION THEOREM. The number of equivalence classes of functions in R D determined by the power group B ~ is </p><p>where </p><p>N(Ba ) _ 1 ~ Z(A; m~(~), m2(fl),..., ma(fi)), </p><p>m~(~)= Y, sjX~). slk </p><p>(12) </p></li><li><p>UNSOLVED GRAPHICAL ENUMERATION PROBLEMS, IV 9 </p><p>As an application of this theorem, we count the number b~ of graphs that are unique up to isomorphism or complementation, so that each graph is equivalent to its complement as well as itself. The group whose orbits are these equivalence classes is the group B s*, where B = S~ 2). Let fix = (1)(2) and fi~ = (1, 2). Then we have </p><p>lO, k odd, mk(fil ) = 2 and mk(fi2) = 2, k even. </p><p>Substituting these values into (12), we have </p><p>N((S(~2))s 0 ---- I[Z(S(~); 2, 2,...) + Z(v(-~); 0, 2,...)1. </p><p>From this result one can easily obtain a formula for the number of self-complementary graphs [12]. Indeed, the number c, of self-complemen- tary graphs is </p><p>% ---- 2b~ -- g~, </p><p>where g~ is the number of graphs with p points, so that </p><p>c~ = Z(S~2); 0, 2, 0, 2,...). (13) </p><p>The original formulation of this enumeration theorem by de Bruijn [I], although considerably different in appearance, is equivalent to the Power Group Enumeration Theorem. In particular, any problem that can be solved by his method can also be solved by the formula in Harary and Palmer [9] which gives the variable form of (12), just as P61ya's Theorem (7) is the variable form of Burnside's Theorem (1). </p><p>4. UNSOLVED GRAPHICAL ENUMERATI...</p></li></ul>